Question Number 55970 by necx1 last updated on 07/Mar/19
![](https://www.tinkutara.com/question/7448.png)
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Mar/19
![10μF and 10μF are parallel so 10+10=20μF 15μF and 5μF are parallel so 15+5=20μF now this 20μF and 20μF are series c_(eq) =(1/((1/(20))+(1/(20))))=((20)/2)=10μF now this 10μF and 20μF are parallel so 10+20=30μF Q charge=30×10^(−6) ×10=300×10^(−6) Columb... C_(15) :C_5 :C_(20) =15:5:20=3:1:4 so i think charge on 5μF is=(1/(3+1+4))×300×10^(−6) =37.5×10^(−6) C Pls comment and rectify if i am wrong...](https://www.tinkutara.com/question/Q55972.png)
$$\mathrm{10}\mu{F}\:{and}\:\mathrm{10}\mu{F}\:{are}\:{parallel} \\ $$$${so}\:\mathrm{10}+\mathrm{10}=\mathrm{20}\mu{F} \\ $$$$\mathrm{15}\mu{F}\:{and}\:\mathrm{5}\mu{F}\:{are}\:{parallel} \\ $$$${so}\:\mathrm{15}+\mathrm{5}=\mathrm{20}\mu{F} \\ $$$${now}\:{this}\:\mathrm{20}\mu{F}\:{and}\:\mathrm{20}\mu{F}\:{are}\:{series} \\ $$$${c}_{{eq}} =\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{20}}+\frac{\mathrm{1}}{\mathrm{20}}}=\frac{\mathrm{20}}{\mathrm{2}}=\mathrm{10}\mu{F} \\ $$$${now}\:{this}\:\mathrm{10}\mu{F}\:{and}\:\mathrm{20}\mu{F}\:{are}\:{parallel} \\ $$$${so}\:\mathrm{10}+\mathrm{20}=\mathrm{30}\mu{F} \\ $$$${Q}\:{charge}=\mathrm{30}×\mathrm{10}^{−\mathrm{6}} ×\mathrm{10}=\mathrm{300}×\mathrm{10}^{−\mathrm{6}} {Columb}… \\ $$$${C}_{\mathrm{15}} :{C}_{\mathrm{5}} :{C}_{\mathrm{20}} =\mathrm{15}:\mathrm{5}:\mathrm{20}=\mathrm{3}:\mathrm{1}:\mathrm{4} \\ $$$${so}\:{i}\:{think}\:{charge}\:{on}\:\mathrm{5}\mu{F} \\ $$$${is}=\frac{\mathrm{1}}{\mathrm{3}+\mathrm{1}+\mathrm{4}}×\mathrm{300}×\mathrm{10}^{−\mathrm{6}} =\mathrm{37}.\mathrm{5}×\mathrm{10}^{−\mathrm{6}} {C} \\ $$$$\boldsymbol{{P}}{ls}\:{comment}\:{and}\:{rectify}\:{if}\:{i}\:{am}\:{wrong}… \\ $$$$ \\ $$